I'm struggling on finding what's wrong with my model of the following system which seems to break the second law:

Imagine two punctual perfect black bodies at two temperature $T1 $ and $T2$. Each black body is perfectly isolated at the center of a spherical mirror which reflect all the emitted radiation on it. In each spherical mirror, there is a hole defined by the difference between the sphere and a cone of angle $a1 $ or $a2 $ whose summit is on the corresponding black body. The 2 hole face each other, and lengths duct all the rays escaping from one black body to the other one:

scheme of the system

At the equilibrium, the power exchange between the two black bodies are the same:

$P1 = P2$

the Stefan–Boltzmann law gives us:

$ \sigma T1^4 S1 = \sigma T2^4.S2$ with S1 and S2 the surface of the spherical hole

$ T1^4.R²/2*a1 = T2^4.R²/2*a2$ with $R $ the radius of the spherical mirrors

$ T1/T2 = (a2/a1)^{1/4}$ $T1 = k.T2 $ with k in R+

This result shows that with the good set of value for $a1 $ and $a2$, the system can reach an equilibrium for any (relative) values for $T1 $ and $T2$. This is really disturbing for me. For instance I could have a configuration where at beginning T1=T2 and at the equilibrium $T1>>T2$, which seems to break the second thermodynamic law... (if this is real, I put a thermal engin between the two black bodies to generate unlimited power ^^)

What's wrong with this approach? I'm suspecting that the Stefan–Boltzmann law doesn't apply like that if the black body already receive radiations, but I have no clue about this


2 Answers 2


The Stefan Boltzmann law is the wrong law to use for this problem. The Stefan Boltzmann law describes the total power radiated by a black body, not the power transferred between two black bodies. The correct law is the radiative heat transfer law which can be derived from the Stefan Boltzmann law restricted appropriately for the geometry: $$\dot Q_{1\rightarrow 2}=\sigma A_1 F_{1\rightarrow 2}(T_1^4-T_2^4)$$ where the main thing you were missing in your expression is $F_{1\rightarrow 2}$ which is the view factor from object 1 to object 2.

Since $A_1 F_{1\rightarrow 2}=A_2 F_{2\rightarrow 1}$ it doesn’t matter which object you consider. You only get $\dot Q=0$ for $T_1=T_2$

  • $\begingroup$ Thank you, However I imagine that in my case A1F1→2 would be equivalent to considering that the outcoming power is a fraction of the total emitted power, with this fraction equal to the spherical surface of the hole divided by the total surface of the sphere. Intuitively it seems inline with the assumption that all emitted ray are evenly distributed around the punctual black body, and thus only those going through the hole contribute to the exchange of power. If I'm wrong with this, what is the physical interpretation of the escaping power not proportional to the escaping area ratio? $\endgroup$ Dec 23, 2020 at 11:29
  • $\begingroup$ @vaferdolosa $A_1$ is the area of object 1 that is “exposed” to object 2 and $F_{1\rightarrow 2}$ is the fraction of rays leaving $A_1$ that reach $A_2$. Both the area and the view factor are important. Together we see that every ray from object 1 to object 2 there is a ray from object 2 to object 1. So the product is the same either way. You appear to be considering only $A$ and neglecting $F$ $\endgroup$
    – Dale
    Dec 23, 2020 at 13:00
  • $\begingroup$ I understand, but I have difficulties with "for every ray from 1 to 2 there is a ray from 2 to 1". Of course this is continuous (meaning infinite number of ray) so in some extent it could be mathematically true, but if I we consider that we can count these rays (meaning there is a finite number of them), the system with the larger opening sends more rays than the other one. Imagine a1=45° and a2=1°, in my understanding A1 >> A2 while F1→2 = F2→1 = 1 (thanks to the lenses, no ray is "lost") . Where am I wrong? Thank you for your patience with me ^^ $\endgroup$ Dec 23, 2020 at 13:41
  • $\begingroup$ @vaferdolosa Your “thanks to the lenses no ray is lost” is physically impossible. There is no possible configuration of mirrors or lenses that will make that correct. Since $A_1/A_2=45^2$ then $F_{1\rightarrow 2}/F_{2\rightarrow 1}=1/45^2$ for any configuration of lenses. This is the conservation of etendue. It is true that “the system with the larger opening sends more rays than the other one” but not all of those rays sent by the large one will reach the small one. $\endgroup$
    – Dale
    Dec 23, 2020 at 14:03
  • $\begingroup$ That's a good point, however I may be wrong but I think I'm able to propose such a configuration: At each output put a lens which focal point is the black body. Then all your ray are parallel and contained in cylinders of radius Ri given by tan(ai) = Ri/fi => fi = Ri/tan(ai). Knowing that you can chose two lenses with focalsf1 anf f2 to get the same radius for both ray cylinders of the two systems, align them and each outcoming ray of a system will be focused on the other black body. (I don't know if it is clear without schema?) Where is the mistake in my solution? Once again thank you $\endgroup$ Dec 23, 2020 at 16:35

Here is a more detailed shema, especially with the optical peering system detailed:

enter image description here

Lets call A the area of a black body (the 2 are identical), reciprocity rule gives us:

$A.F_{1\rightarrow 2} = A.F_{2\rightarrow 1}$


$F_{1\rightarrow 2} + F_{1\rightarrow M1} = 1$

$F_{2\rightarrow 1} + F_{2\rightarrow M2} = 1$

After substituation:

$F_{1\rightarrow M1} = F_{2\rightarrow M2}$

So, as the two black bodies have the same area, this would means the two mirrors receive the same amount of incoming power from their black body, whereas their surface exposition are largely different from each other, while the configuration is the same.

This shows there is a flaw somewhere. It highlights that their must be another power source/drain for our mirror explaining this equality.

For me the issue comes from the ponctual sources. If they have an emitting outter area, this means they are not punctual (as pointed by @Dale). This implies that the image of one black body is not perfectly the image of the other one (bigger or smaller). Thus, some rays may miss the pseudo punctual black body, reaching the mirror and probably coming back to the emitter, which explains the previous results and why the modelisation of the system is wrong (can't use punctual source here)

Thank you @Dale for your help, working with view factors helped me to figure out this

  • $\begingroup$ In my opinion the other answer is complete and should have been accepted. $\endgroup$
    – boyfarrell
    Dec 26, 2020 at 21:42
  • $\begingroup$ The other answer was relevant, but didn't point to the source of the issue (which was my question) i.e. the use of a punctual source for this modelisation, which is incompatible with black body radiation model. Wether you formalize this problem with ray tracing or view factors doesn't change anything. However the use of view factor was helpfull as it highlighted unexpected consequences in the power balance, as detailed in this answer. $\endgroup$ Dec 27, 2020 at 15:44

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